Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;

import static java.lang.Math.*;
import static java.util.Arrays.*;

public class ProblemC3 {
    BufferedReader rd;

    ProblemC3() throws IOException {
        rd = new BufferedReader(new InputStreamReader(System.in));
        compute();
    }

    private void compute() throws IOException {
        int n = intarr()[0];
        int[][] c = new int[n][];
        for(int i=0;i<n;i++) {
            c[i] = intarr();
        }

        List<int[]> res1 = solve(deepcopy(c), true);
        if(res1 == null) {
            out(-1);
        } else {
            List<int[]> res2 = solve(deepcopy(c), false);
            List<int[]> res = computeMoveCount(res1) < computeMoveCount(res2)?res1:res2;

            int moveCount = computeMoveCount(res);
            StringBuilder buf = new StringBuilder();
            buf.append(moveCount);
            for (int[] r : res) {
                for (int i = 0; i < r[1]; i++) {
                    buf.append('\n').append(r[0] == 0 ? "row" : "col").append(' ').append(r[2]+1);
                }
            }
            out(buf);
        }
    }

    private int computeMoveCount(List<int[]> x) {
        int moveCount = 0;
        for (int[] r : x) {
            moveCount += r[1];
        }
        return moveCount;
    }

    private int[][] deepcopy(int[][] c) {
        int[][] res = new int[c.length][];
        for(int i=0;i<c.length;i++) {
            res[i] = copyOf(c[i], c[i].length);
        }
        return res;
    }

    private List<int[]> solve(int[][] c, boolean flip) {
        int n = c.length;
        int m = c[0].length;
        List<int[]> res = new ArrayList<>();
        for(int k=0;k<=2;k++) {
            if(k==0 ^ flip) for (int i = 0; i < n; i++) {
                int mi = 1000;
                for (int j = 0; j < m; j++) {
                    mi = min(mi, c[i][j]);
                }
                if (mi > 0) {
                    res.add(new int[]{0, mi, i});
                    for (int j = 0; j < m; j++) {
                        c[i][j] -= mi;
                    }
                }
            }
            if(k==1 ^ flip) for (int i = 0; i < m; i++) {
                int mi = 1000;
                for (int j = 0; j < n; j++) {
                    mi = min(mi, c[j][i]);
                }
                if (mi > 0) {
                    res.add(new int[]{1, mi, i});
                    for (int j = 0; j < n; j++) {
                        c[j][i] -= mi;
                    }
                }
            }
        }
        for(int i=0;i<n;i++) {
            for(int j=0;j<m;j++) {
                if(c[i][j] > 0) {
                    return null;
                }
            }
        }
        return res;
    }

    private int[] intarr() throws IOException {
        return intarr(rd.readLine());
    }

    private int[] intarr(String s) {
        String[] q = split(s);
        int n = q.length;
        int[] a = new int[n];
        for(int i=0;i<n;i++) {
            a[i] = Integer.parseInt(q[i]);
        }
        return a;
    }

    private String[] split(String s) {
        if(s == null) {
            return new String[0];
        }
        int n = s.length();
        int start = -1;
        int end = 0;
        int sp = 0;
        boolean lastWhitespace = true;
        for(int i=0;i<n;i++) {
            char c = s.charAt(i);
            if(isWhitespace(c)) {
                lastWhitespace = true;
            } else {
                if(lastWhitespace) {
                    sp++;
                }
                if(start == -1) {
                    start = i;
                }
                end = i;
                lastWhitespace = false;
            }
        }
        if(start == -1) {
            return new String[0];
        }
        String[] res = new String[sp];
        int last = start;
        int x = 0;
        lastWhitespace = true;
        for(int i=start;i<=end;i++) {
            char c = s.charAt(i);
            boolean w = isWhitespace(c);
            if(w && !lastWhitespace) {
                res[x++] = s.substring(last,i);
            } else if(!w && lastWhitespace) {
                last = i;
            }
            lastWhitespace = w;
        }
        res[x] = s.substring(last,end+1);
        return res;
    }

    private boolean isWhitespace(char c) {
        return c==' ' || c=='\t';
    }

    private static void out(Object x) {
        System.out.println(x);
    }

    public static void main(String[] args) throws IOException {
        new ProblemC3();
    }
}